Include from another URL?

Hi there,
I'm new to these groups so forgive me if I'm asking at the wrong place
(tell me where then :)

Ok,
I'm not PHP guru but I need to create a simple script that would do
this:

Include a content from another URL into the current output. I have
done something like
this:

------ code start
<?php
include 'http://myotherurl.com:8080';
?>
------ code end

and it seems to work with one exception: it produces a warning message
saying something like
"Warning: main(): stream does not support seeking in
..../web-root/index.php on line 9"
where line #9 is: include 'http://myotherurl.com:8080'; from above
code.
It then shows a message from 'http://myotherurl.com:8080'

Thanks!!!
Nik

On Thu, 06 May 2004 13:00:56 -0700, x user wrote:

> Hi there,
> I'm new to these groups so forgive me if I'm asking at the wrong place
> (tell me where then :)
>
> Ok,
> I'm not PHP guru but I need to create a simple script that would do
> this:
>
> Include a content from another URL into the current output. I have
> done something like
> this:
>
> ------ code start
> <?php
> include 'http://myotherurl.com:8080';
> ?>
> ------ code end
>
> and it seems to work with one exception: it produces a warning message
> saying something like
> "Warning: main(): stream does not support seeking in
> .../web-root/index.php on line 9"
> where line #9 is: include 'http://myotherurl.com:8080'; from above
> code.
> It then shows a message from 'http://myotherurl.com:8080'
>
> Thanks!!!
> Nik

Hi Nik,
try this, just replace the php.net URL with the one you want:

<?php
$file = fopen ("http://www.php.net/", "r");
if (!$file) {
echo "<p>Unable to open remote file.\n";
exit;
}
while (!feof ($file)) {
$line = fgets ($file, 1024);
print $line;
}
fclose($file);
?>

I think the problem with include is that PHP tries to interpret what you
are including, instead of displaying it.

cheers,
DrTebi